In one year there was 675 deaths due to impaired driving. For any randomly selected day in one year what is the probability of 2 impaired driving related deaths?
To find the probability of 2 impaired driving related deaths on any randomly selected day in one year, we need to use the concept of probability and the given information.
Given:
- Total number of deaths due to impaired driving in one year = 675
To calculate the probability, we need to know the average number of impaired driving related deaths per day. We can find this by dividing the total number of deaths (675) by the number of days in a year.
Let's assume there are 365 days in a year (excluding leap years):
Average number of impaired driving related deaths per day = 675 / 365
Once we have the average number of impaired driving related deaths per day, we can use the Poisson distribution to calculate the probability of having 2 deaths on a randomly selected day.
The Poisson distribution formula is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- P(X = k) is the probability of having k events occur
- λ (lambda) is the average number of events occurring in a given time period
- k is the number of events we are interested in
For this case, k = 2 (since we want to find the probability of 2 impaired driving related deaths), and λ = Average number of impaired driving related deaths per day.
Now, let's plug in the values into the formula:
P(X = 2) = (e^(-λ) * λ^2) / 2!
Calculating this value will give us the probability of having 2 impaired driving related deaths on any randomly selected day in one year.